Question

use the shell method to write and evaluate the definite integral that represents the volume of the solid generated by revolving the plane region about the x-axis.
$y = \sqrt{x + 30}$
$y = x$
$y = 0$

Explanation:

Step1: Find intersection points

First, find where $y = x$ and $y = \sqrt{x+30}$ intersect:
Set $x = \sqrt{x+30}$, square both sides: $x^2 = x + 30$
Rearrange: $x^2 - x - 30 = 0$
Factor: $(x-6)(x+5)=0$, so $x=6$ (since $x=-5$ gives $y=-5$, but $y=\sqrt{x+30}\geq0$, so valid intersection is $(6,6)$)
Find where $y=\sqrt{x+30}$ meets $y=0$: $0=\sqrt{x+30} \implies x=-30$, so point $(-30,0)$
Find where $y=x$ meets $y=0$: $(0,0)$

For shell method about x-axis, we use $y$ as variable. Rewrite equations in terms of $x$:
$y = \sqrt{x+30} \implies x = y^2 - 30$
$y = x \implies x = y$
The range of $y$ is $0 \leq y \leq 6$.

Step2: Set up shell volume integral

Shell radius = $y$, shell height = right function - left function = $y - (y^2 - 30) = -y^2 + y + 30$
Volume formula for shell method about x-axis:
$$V = 2\pi \int_{a}^{b} (\text{radius})(\text{height}) dy$$
Substitute bounds and functions:
$$V = 2\pi \int_{0}^{6} y(-y^2 + y + 30) dy = 2\pi \int_{0}^{6} (-y^3 + y^2 + 30y) dy$$

Step3: Evaluate the integral

Find antiderivative:
$$\int (-y^3 + y^2 + 30y) dy = -\frac{y^4}{4} + \frac{y^3}{3} + 15y^2 + C$$
Evaluate from 0 to 6:
At $y=6$:
$-\frac{6^4}{4} + \frac{6^3}{3} + 15(6^2) = -\frac{1296}{4} + \frac{216}{3} + 15(36) = -324 + 72 + 540 = 288$
At $y=0$, the antiderivative is 0.
So the definite integral equals $288 - 0 = 288$

Step4: Compute final volume

Multiply by $2\pi$:
$V = 2\pi \times 288$

Answer:

$576\pi$